A pitcher throws a 0.5kg ball of clay at a 6kg block of wood. The clay sticks to the wood on impact and their joint velocity afterwards is 3m/s. What is the original speed of the clay? I know the answer is 39m/s but I have no idea how they come up with that
39m/s
To find the original speed of the clay, we can use the law of conservation of momentum.
According to the law of conservation of momentum, the momentum before the collision is equal to the momentum after the collision. Mathematically, this can be expressed as:
(mass of object 1 * velocity of object 1) + (mass of object 2 * velocity of object 2) = (total mass * final velocity)
Let's denote the mass of the clay as m1, the velocity of the clay as v1, the mass of the wood as m2 (6kg), and the final velocity as vf (3m/s). We can also assume that the wood is initially at rest, so its velocity (v2) is 0.
Plugging the given values into the equation, we get:
(m1 * v1) + (m2 * v2) = (m1 + m2) * vf
Since v2 is 0, the equation simplifies to:
(m1 * v1) = (m1 + m2) * vf
Now, let's substitute the known values:
(0.5kg * v1) = (0.5kg + 6kg) * 3m/s
Now we can solve for v1:
0.5kg * v1 = 6.5kg * 3m/s
0.5kg * v1 = 19.5kg m/s
v1 = (19.5kg m/s) / 0.5kg
v1 = 39m/s
So the original speed of the clay is 39m/s.
To find the original speed of the clay before impact, we can use the principle of conservation of momentum.
The principle of conservation of momentum states that the total momentum of a system of objects remains constant if no external forces act on it. In other words, the initial momentum of the system should be equal to the final momentum of the system.
Let's denote the original speed of the clay as v, and the mass of the block of wood as m1 (6kg) and the mass of the clay as m2 (0.5kg).
The initial momentum of the system is given by:
Initial momentum = (mass of clay) x (velocity of clay) + (mass of wood) x (velocity of wood)
Final momentum of the system is given by:
Final momentum = (mass of clay + mass of wood) x (final velocity of clay and wood)
According to the conservation of momentum, the initial momentum is equal to the final momentum:
(m2 x v) + (m1 x 0) = (m1 + m2) x 3
Since the wood is initially at rest (velocity = 0), the momentum of the wood is zero.
Now, let's plug in the given values:
(0.5kg x v) + (6kg x 0) = (6kg + 0.5kg) x 3
0.5v = 18.5
Dividing both sides by 0.5:
v = 18.5 / 0.5
Therefore, the original speed of the clay is:
v = 37 m/s
It seems there is an error in the given answer. The correct answer is 37 m/s, not 39 m/s.
So, the original speed of the clay is 37 m/s, not 39 m/s.
momentum is conserved
the mass of the clay, times its velocity; is equal to the mass of the clay+block, times its velocity
0.5 * v = (6 + 0.5) * 3